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4 years ago

Can someone plz help me solved this problem! I’m giving you 10 points! I need help plz help me! Will mark you as brainiest!

Mathematics

1 answer:

pickupchik [31]4 years ago

7 0

Hey there! :)

Answer:

a. 3

b. -22

c. -2

d. -2

e. 5a + 8

f. a² + 6a + 3

Step-by-step explanation:

Calculate the answers by substituting the values inside of the parenthesis for 'x':

a. f(1) = 5(1) - 2 = 3

b. f(-4) = 5(-4) - 2 = -22

c. g(-3) = (-3)² + 2(-3) - 5 = 9 - 6 - 5 = -2

d. g(1) = 1² + 2(1) - 5 = 1 + 2 -5 = -2

e. f(a+ 2) = 5(a+2) - 2 = 5a + 10 - 2 = 5a + 8

f. g(a + 2) = (a + 2)² + 2(a + 2) - 5 = a² + 4a + 4 + 2a + 4 - 5 =

a² + 6a + 3

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Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 229 wi

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Answer:

A. 99% CI for p: 0.125 < p < 0.259

B. 99% CI for p: 0.807 < p < 0.911

C. 95% CI for p: 0.776 < p < 0.844

D. 95% CI using t-distribution: 23.7 < µ < 40.3

Step-by-step explanation:

$p=\frac{x}{n} =\frac{44}{229} = 0.192139738$

Confidence = 99% = 0.99

$\alpha$ =1 - confidence = 1 - 0.99 = 0.01

Critical z-value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.192139738(1-0.192139738)}{229} }$

$SE=\sqrt{\frac{0.155222059}{229} }$

SE = 0.026035

E = Margin of Error

E = z-value * SE = 2.58 * 0.026035

E = 0.067170516

Lower Limit = p - E = 0.192139738 - 0.067170516 = 0.125

Upper Limit = p + E = 0.192139738 + 0.067170516 = 0.259

99% CI for p: 0.125 < p < 0.259

$p=\frac{x}{n} =\frac{256}{298} = 0.8590604$

Confidence = 99% = 0.99

$\alpha$ =1 - confidence = 1 - 0.99 = 0.01

Critical z-value = $z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.8590604(1-0.8590604)}{298} }$

$SE=\sqrt{\frac{0.121075629}{298} }$

SE = 0.020157

E = Margin of Error

E = z-value * SE = 2.58 * 0.020157

E = 0.052004

Lower Limit = p - E = 0.8590604 - 0.052004 = 0.807

Upper Limit = p + E = 0.8590604 + 0.052004 = 0.911

99% CI for p: 0.125 < p < 0.259

$p=\frac{x}{n} =\frac{405}{500} = 0.81$

Confidence = 95% = 0.95

$\alpha$ =1 - confidence = 1 - 0.95 = 0.05

Critical z-value = $z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$ (from z-table)

SE = Standard Error of p

$SE=\sqrt{\frac{p(1-p)}{n} }$

$SE=\sqrt{\frac{0.81(1-0.81)}{500} }$

$SE=\sqrt{\frac{0.1539}{500} }$

SE = 0.017544

E = Margin of Error

E = z-value * SE = 1.96 * 0.017544

E = 0.034387

Lower Limit = p - E = 0.81 - 0.034387 = 0.776

Upper Limit = p + E = 0.81 + 0.034387 = 0.844

95% CI for p: 0.776 < p < 0.844

x = sample mean = 32

s = sample standard deviation = 13

n = sample size = 12

Confidence = 95% = 0.95

$\alpha$ =1 - confidence = 1 - 0.95 = 0.05

Being the sample size less than 30, we use the statistical t.

df = degree of freedom = n - 1 = 12 - 1 = 11

Critical t-value = $t_{\alpha/2,df}=t_{0.05/2,11}=t_{0.025,11}=2.201$ (from t-table, two tails, df=11)

SE = standard error

$SE=\frac{s_{x} }{\sqrt{n} } =\frac{13 }{\sqrt{12} }=3.752777$

E = margin of error

E = t-value * SE = 2.201 * 3.752777 = 8.259862

Lower Limit = x - E = 32 - 8.259862 = 23.7

Upper Limit = x + E = 32 + 8.259862 = 40.3

95% CI using t-distribution: 23.7 < µ < 40.3

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